# What's the condition for a plane and a line to be coplanar in 3D?

Please correct me if I'm wrong, but given a plane expressed in point-normal form, and a line expressed in parametric form, an easy way of finding the intersection (or lack of it) is substituting the $$(x, y, z)$$ values given by the line equation into the equation of the plane. If a real value is found for the parameter, that means the line and the plane intersect on a single point, which can be found by substituting the value found for the parameter into the equation of the line. If when intersecting the line and plane we reach a result such as $$1=0$$, we know the line does not intersect the plane. So far so good. But... what would happen if the plane and the line are coplanar? That is, there are infinite solutions to their intersection?

Examples:

i. $$\pi) \ x +y+z=3, \ r)\left\{\begin{matrix} x=3\lambda & \\ y=2\lambda & \\ z=\lambda \end{matrix}\right.$$

Substituting gives $$\lambda = 0.5$$

Thus $$\ pi \cap r = P(1.5,1,0.5)$$

ii. $$\pi) \ x +y+z=3, \ r)\left\{\begin{matrix} x=-0.5\lambda & \\ y=-0.5\lambda & \\ z=\lambda \end{matrix}\right.$$

Substituting gives $$0=3$$

Thus $$\ pi \cap r = \varnothing$$

• If you have a plane $ax+by+cz=d$ with $u=(a,b,c)$ and the line is defined as parallel to $u’=(a’,b’,c’)$ there are three cases: 1) $u\times{}u’=0$ and either a) the plane contains the line or b) the line and plane are disjoint 2) $u\times{}u’\ne{}0$ and you have just one point of intersection. – Μάρκος Καραμέρης Apr 11 '19 at 18:59
• @Carlos Gruss The minimum distance between any point in the line and the plane is constant. – Narasimham Apr 11 '19 at 21:03
• @Narasimham I like that condition, but (excuse my ignorance), how would you write mathematically the minimum distance between any point in the line and the plane? – Carlos Gruss Apr 11 '19 at 22:10
• Vectors representing a general line and a vector connecting a point on it to reference point should have zero dot product. – Narasimham Apr 11 '19 at 23:43

If the line given parametrically by $$\mathbf p_0+t\mathbf v$$ lies on the plane $$ax+by+cz=\mathbf n\cdot\mathbf p=d$$ then $$\mathbf p_0$$ must satisfy the equation of the plane, i.e., $$\mathbf n\cdot\mathbf p_0=d$$, and its direction vector $$\mathbf v$$ must be perpendicular to the plane’s normal $$\mathbf n$$, so that $$\mathbf n\cdot\mathbf v=0$$. Therefore, when you substitute into the plane’s equation, you will get $$\mathbf n\cdot(\mathbf p_0+t\mathbf v) = d \\ \mathbf n\cdot\mathbf p_0 + t(\mathbf n\cdot\mathbf v) = d \\ d = d.$$

Consider $$\pi) \ x +y+z=3, \ r)\left\{\begin{matrix} x=3\lambda & \\ y=2\lambda & \\ z=3-5\lambda \end{matrix}\right.$$

You will get $$3=3$$ for every $$\lambda$$.

• OK, I see now, thank you. So in this case, $\pi \cap r = r$, right? – Carlos Gruss Apr 11 '19 at 19:08
• Correct. That means every point of the line is also a point on the plane. – Mohammad Riazi-Kermani Apr 11 '19 at 19:09

Given a general line and plane as such:

$$\pi) \ ax +by+cz=d, \\ r)\left\{\begin{matrix} x=p_1+v_1\lambda & \\ y=p_2+v_2\lambda & \\ z=p_3+v_3\lambda \end{matrix}\right.$$

There are two possible cases for the line and plane to be coplanar:

i. $$(ap_1 +bp_2 + cp_3) = 0\ \$$, $$(av_1 +bv_2 + cv_3) = 0\ \$$ and $$\ d = 0$$

ii. $$(ap_1 +bp_2 + cp_3) \ne 0\ \$$, $$(d - ap_1 -bp_2 - cp_3) = 0\ \$$ and $$(av_1 +bv_2 + cv_3) = 0\ \$$

In both cases, $$\pi \cap r = r$$.